please help!!!! how do I find three times the first of two consecutive integers is at least twelve more than four times the second. And to find the greatest possible integers.
Two consecutive integers: x, and x+1
3x > 4(x+1) + 12
3x > 4x + 16
-x > 16
x < -16
So choices would be -16, -15 ... -17, -16 .... -18, -17 ... etc
there are no positive integers that work.
To solve this problem, let's break it down into steps:
Step 1: Set up the equations
Let's assume the first consecutive integer is represented by x, and the second consecutive integer is represented by (x+1). Now we can translate the given statement into an equation:
3x ≥ 4(x+1) + 12
Step 2: Simplify the equation
Let's simplify the equation by distributing the 4 to (x+1):
3x ≥ 4x + 4 + 12
Step 3: Continue simplifying
Combine like terms:
3x ≥ 4x + 16
Step 4: Isolate the variable
To isolate the variable on one side of the inequality, we need to subtract 4x from both sides:
3x - 4x ≥ 16
Step 5: Simplify and solve for x
Combine like terms:
-x ≥ 16
Now, we need to multiply both sides of the inequality by -1. However, we need to remember that when we multiply or divide an inequality by a negative number, we must flip the inequality sign. So, let's multiply both sides by -1 and reverse the inequality sign:
x ≤ -16
Step 6: Interpret the solution
The solution tells us that x, the first consecutive integer, is less than or equal to -16. However, we are looking for the greatest possible integers.
Since we are dealing with consecutive integers, we take the second integer to be (x+1). From our solution, we know that x ≤ -16. Therefore, the greatest possible first integer will be -16, and the second integer will be -16 + 1 = -15.
So, the greatest possible consecutive integers that satisfy the given conditions are -16 and -15.